Answered: Find the value of the chi-square…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

The Fish and Game Department stocked a lake with fish in the following proportions: 30% catfish, 15% bass, 40% bluegill, and 15% pike. Five years later it sampled the lake to see if the distribution of fish had changed. It found that the 500 fish in the sample were distributed as follows.

Catfish	Bass	Bluegill	Pike
122	75	222	81

In the 5-year interval, did the distribution of fish change at the 0.05 level

Find the value of the chi-square statistic for the sample.

Estimate the P-value of the sample test statistic.

P-value > 0.100

0.050 < P-value < 0.100

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

P-value < 0.005

Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories?

Since the P-value > α, we fail to reject the null hypothesis.

Since the P-value > α, we reject the null hypothesis.

Since the P-value ≤ α, we reject the null hypothesis.

Since the P-value ≤ α, we fail to reject the null hypothesis.

Interpret your conclusion in the context of the application.

At the 5% level of significance, the evidence is insufficient to conclude that current fish distribution is different than that of five years ago.

At the 5% level of significance, the evidence is sufficient to conclude that current fish distribution is different than that of five years ago.

Expert Solution

Step 1

The clam of the test is the distribution of fish change or not.

Null hypothesis:

$H_{0} :$ The distribution of fish same.

Alternative hypothesis:

$H_{1} :$ The distribution of fish different.

The expected values are,

Type of Fish	Percent	Observed	Expected
Catfish	30%	122	$500 \times 0.30 = 150$
Bass	15%	75	$500 \times 0.15 = 75$
Bluegill	40%	222	$500 \times 0.40 = 200$
Pike	15%	81	$500 \times 0.15 = 75$
Total		500

The value of the chi-square statistic for the sample is,

$\begin{array}{rcl} χ^{2} & = & \sum_{i} \frac{{(O b s e r v e d - E x p e c t e d)}^{2}}{E x p e c t e d} \\ = & [\frac{{(122 - 150)}^{2}}{150} + \frac{{(75 - 75)}^{2}}{75} + \frac{{(222 - 200)}^{2}}{200} + \frac{{(81 - 75)}^{2}}{75}] \\ = & 5.227 + 2.420 + 0.480 \\ = & 8.127 \end{array}$

Thus, the value of the chi-square statistic for the sample is 8.127.

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